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I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

P.S. I think you can do the following. Start with a square tiling. By drawing a different slant line in each square, you can create a tiling with right trapezoids of pairwise distinct bases and angles.

Now, enumerate these trapezoids, $A_nB_nC_nD_n$ with $A_nB_n \parallel C_nD_n$.

Now, pick $E_1 \in A_1D_1$ and $F_1 \in B_1C_1$ such that $E_1F_1 \not\parallel A_1B_1$ and $A_1B_1E_1F_1, E_1F_1C_1D_1$ are afinelly equivalent.

Next, for each $N$ pick, $\phi : A_1B_1C_1D_1 \to A_nB_nC_nD_n$ an affine equivalence. Let $E_nF_n = \phi(E_1F_1)$. Split $A_nB_nC_nD_n$ into $A_nB_nF_nE_n$ and $E_nF_nC_nD_n$.

All the quadrilaterals are not trapezoids, and they are affinely equivalent. Now, making the right choices of distinct bases and angles in the first step, I think you can make sure all these quadrialterals are non-congruent (they all have a $90^\circ$ angle, and you have control on an edge to this angle and the angle at the other end of this edges).

I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

PS Let $ABCD$ be a trapezoid with $AB \parallel CD$. Pick arbitrary points $E \in AD, F \in BC$ and draw $EF$. This allows you to turn any trapezoid tiling into a nontrapezoid tiling.

I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

PS Let $ABCD$ be a trapezoid with $AB \parallel CD$. Pick arbitrary points $E \in AD, F \in BC$ and draw $EF$. This allows you to turn any trapezoid tiling into a nontrapezoid tiling.